$h(n) = 2n^{2}+4n-4+f(n)$ $f(t) = -1$ $g(t) = 7t^{3}-6t^{2}-3t+4+4(h(t))$ $ f(g(2)) = {?} $
First, let's solve for the value of the inner function, $g(2)$ . Then we'll know what to plug into the outer function. $g(2) = 7(2^{3})-6(2^{2})+(-3)(2)+4+4(h(2))$ To solve for the value of $g$ , we need to solve for the value of $h(2)$ $h(2) = 2(2^{2})+(4)(2)-4+f(2)$ To solve for the value of $h$ , we need to solve for the value of $f(2)$ $f(2) = -1$ $f(2) = -1$ That means $h(2) = 2(2^{2})+(4)(2)-4-1$ $h(2) = 11$ That means $g(2) = 7(2^{3})-6(2^{2})+(-3)(2)+4+(4)(11)$ $g(2) = 74$ Now we know that $g(2) = 74$ . Let's solve for $f(g(2))$ , which is $f(74)$ $f(74) = -1$